If there are 10 positive real numbers `n_1 lt n_2 lt n_3 ...... lt n_(10)`. How many triplets of these numbers ` (n_1, n_2, n_3), (n_2, n_3, n_4)`, ..... can be generated such that in each triplet the first number is always less than the second number and the second number is always less than the third number?

Options

A. 45
B. 90
C. 120
D. 180

Answer

To solve the problem of finding how many triplets can be generated from the 10 positive real numbers \( n_1, n_2, n_3, \ldots, n_{10} \) such that each triplet maintains the order \( n_i < n_j < n_k \), we can follow these steps:

### Step 1: Understand the Problem
We need to select triplets of numbers from a set of 10 numbers where the order of selection matters. The triplet must satisfy the condition that the first number is less than the second, and the second is less than the third.

### Step 2: Choose 3 Numbers
From the 10 numbers, we need to choose 3 numbers. The number of ways to choose 3 numbers from 10 can be calculated using the combination formula:

\[ \text{Number of ways to choose 3 from 10} = \binom{10}{3} \]

### Step 3: Calculate the Combination
The formula for combinations is given by:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

For our case, \( n = 10 \) and \( r = 3 \):

\[ \binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \cdot 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 \]

### Step 4: Conclusion
Thus, the total number of triplets \( (n_i, n_j, n_k) \) such that \( n_i < n_j < n_k \) is 120.

### Final Answer
The number of valid triplets is \( \boxed{120} \).

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